Continunity as a property of functions - AP Calculus AB

is not differentiable at and because the values are discontinuities. is not differentiable at because that point is a corner, indicating that the one-side limits at are different. is differentiable:the one side limits are the same and the point is continuous.

The answer is both.

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:

$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.

Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

, and

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

.

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

Note that , indicating that there is a horizontal tangent on at . More specifically, the derivative is the slope of the tangent line. If the slope of the tangent line is 0, then the tangent is horizontal.

The other two are incorrect because sharp turns only apply when we want to take the derivative of something. The derivative of a function at a sharp turn is undefined, meaning the graph of the derivative will be discontinuous at the sharp turn. (To see why, ask yourself if the slope at is positive 1 or negative 1?) On the other hand, integration is less picky than differentiation: We do not need a smooth function to take an integral.

In this case, to get from to , we took an integral, so it didn't matter that there was a sharp turn at the specified point. Thus, neither function had any discontinuities.

Unless we are explicitly told so, via graph, information, or otherwise, we cannot assume is continuous at unless , which is required for to be continuous at .

We cannot assume anything about the existence of , because we do not know what is, or its end behavior.

For a function to be continuous the following criteria must be met:

1. The function must exist at the point (no division by zero, asymptotic behavior, negative logs, or negative radicals).
2. The limit must exist.
3. The point must equal the limit. (Symbolically, ).

It is easiest to first find any points where the function is undefined. Since our function involves a fraction and a natural log, we must find all points in the domain such that the natural log is less than or equal to zero, or points where the denominator is equal to zero.

To find the values that cause the natural log to be negative we set

Therefore, those x values will yield our points of discontinuity. Normally, we would find values where the natural log is negative; however, for all the function is positive.

The answer is both.

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:

$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.

Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

, and

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

.

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

is not differentiable at and because the values are discontinuities. is not differentiable at because that point is a corner, indicating that the one-side limits at are different. is differentiable:the one side limits are the same and the point is continuous.

Note that , indicating that there is a horizontal tangent on at . More specifically, the derivative is the slope of the tangent line. If the slope of the tangent line is 0, then the tangent is horizontal.

The other two are incorrect because sharp turns only apply when we want to take the derivative of something. The derivative of a function at a sharp turn is undefined, meaning the graph of the derivative will be discontinuous at the sharp turn. (To see why, ask yourself if the slope at is positive 1 or negative 1?) On the other hand, integration is less picky than differentiation: We do not need a smooth function to take an integral.

In this case, to get from to , we took an integral, so it didn't matter that there was a sharp turn at the specified point. Thus, neither function had any discontinuities.

Unless we are explicitly told so, via graph, information, or otherwise, we cannot assume is continuous at unless , which is required for to be continuous at .

We cannot assume anything about the existence of , because we do not know what is, or its end behavior.

For a function to be continuous the following criteria must be met:

1. The function must exist at the point (no division by zero, asymptotic behavior, negative logs, or negative radicals).
2. The limit must exist.
3. The point must equal the limit. (Symbolically, ).

It is easiest to first find any points where the function is undefined. Since our function involves a fraction and a natural log, we must find all points in the domain such that the natural log is less than or equal to zero, or points where the denominator is equal to zero.

To find the values that cause the natural log to be negative we set

Therefore, those x values will yield our points of discontinuity. Normally, we would find values where the natural log is negative; however, for all the function is positive.

The answer is both.

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:

$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.

Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

, and

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

.

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

is not differentiable at and because the values are discontinuities. is not differentiable at because that point is a corner, indicating that the one-side limits at are different. is differentiable:the one side limits are the same and the point is continuous.

Note that , indicating that there is a horizontal tangent on at . More specifically, the derivative is the slope of the tangent line. If the slope of the tangent line is 0, then the tangent is horizontal.

The other two are incorrect because sharp turns only apply when we want to take the derivative of something. The derivative of a function at a sharp turn is undefined, meaning the graph of the derivative will be discontinuous at the sharp turn. (To see why, ask yourself if the slope at is positive 1 or negative 1?) On the other hand, integration is less picky than differentiation: We do not need a smooth function to take an integral.

In this case, to get from to , we took an integral, so it didn't matter that there was a sharp turn at the specified point. Thus, neither function had any discontinuities.

Unless we are explicitly told so, via graph, information, or otherwise, we cannot assume is continuous at unless , which is required for to be continuous at .

We cannot assume anything about the existence of , because we do not know what is, or its end behavior.